An AC source of angular frequency ω is connected across a resistor R and a capacitor C in series. The current is observed as I. Now the frequency is changed to ω/4 (keeping the voltage constant). The current is found to be I/3. Find the ratio of resistance to reactance at frequency ω.

In a series RC circuit, the impedance is given by:

Z = √(R² + X_C²)

where R is the resistance and X_C is the capacitive reactance. The current flowing through the circuit is:

I = V/Z

At angular frequency ω, let the capacitive reactance be X. Therefore:

Z₁ = √(R² + X²)

Since the current is I:

I = V / √(R² + X²)

When the frequency becomes ω/4, the capacitive reactance becomes:

X' = 1/(ω/4)C = 4X

The new impedance becomes:

Z₂ = √(R² + (4X)²)

The new current is given as I/3:

I/3 = V / √(R² + 16X²)

Dividing the two current equations:

(V/√(R² + X²)) ÷ (V/√(R² + 16X²)) = 3

√(R² + 16X²) = 3√(R² + X²)

Squaring both sides:

R² + 16X² = 9(R² + X²)

R² + 16X² = 9R² + 9X²

7X² = 8R²

Therefore:

R²/X² = 7/8

Taking square root:

R/X = √(7/8)

Hence, the ratio of resistance to capacitive reactance at frequency ω is:

R : X_C = √(7/8)

Final Answer

√(7/8)